Chapter – One

1. Special theory of Relativity

In physics special theory of relativity is generally accepted and experimentally well – confirmed physical
Theory regarding the relation ship b/n space and time . All motion is relative; the seed of light in free
space is the same for all observers.

1.1 Frame of Reference

An inertial frame of reference is one in which Newton’s 1st law of motion holds. In such frame, an object
at rest remains at rest and an object in motion continues to move at constant velocity if no force act on
it. All inertial frames are equally valid.

There is no universal frame of reference that can be used every where, no such things us “absolute
motion.” The theory of relativity deals with the consequence of the lack of a universal frame of
reference. Special theory of relativity, which is what Einstein published in 1905 treats problem that
involve inertial frame of reference.

1.2 Galilian Transformation.

Before special relativity, transforming measurements from one inertial system to an other seemed
obvious . If clocks in both systems are started when the origins of S and S' coincide, measurements in
the x-direction made in ‘’S’’ will be greater than those made in S’ by the a mount ‘’vt” which is the
distance S’ has moved in the x-direction. That is : x’ = x – vt……………………...(1.1)

There is no relative motion in the y and z directions and so

y’ = y …………………………….(1.2)

Consider the following diagram with S and S’ of events in x, y, z coordinates.

F.g (1.1) frame S’ moves in the +x direction with the speed of v relative to frame S.

1
The Lorentz transformation must be used to convert made in one of these frames to their equivalents in
the other. i.e z’ = z……………………………..(1.3)

In the absence of any in diction to the contrary in our every day experience, we further assumed that:

t’ = t………………………………….(1.4)

The set of equation (1.1) to equation (1.4) is known as the Galilian transformations for motion. To
convert velocity components measured in the S’ frame according to the Galilian transformation, we
simply differentiate x’, y’, and z’ with respect to time. i.e:

dx '
Vx’ = = vx – v………………………………(1.5)
dt '

dy '
Vy’ = = vy………………………………… (1.6)
dt '

1.3 Postulates of Special Relativity

We have two postulates underlie special theory of relativity. Those are:

 The law of physics are the same in all inertial frames of reference
 Speed of light in free space has the same value in all inertial frames of reference.
1.4 Lorentz Transformation

A reasonable guess a bout the nature of the correct relation ship b/n x and x’ is:

x’ = Y(x – vt)……………………………………….(1.8)

Here Y is a relativity factor that does not depend up on either x or t but may be a function of v . The
choice of equation (1.8) follows from several considerations:

 It is linear in x and x’, so that a single event in frame S corresponds to S’ .

 It is simple, and a simple solution to a problem should always be explored first .
 It has a possibility of reducing to equation (1.1) .

By taking the inverse of equation (1.8) one can write:

x = Y(x’ + vt’)…………………………………………(1.9)

In Galilian transformation, the corresponding coordinates y,y’ and z, z’ which are perpendicular to the
direction of v. Hence we a gain take

y’ = y……………………………………………………….(1.10)

z’ = z……………………………………………………….(1.11)

2
The time coordinates t and t’, how ever are not equal. We can see this by substituting the value of x’
given by equation (1.8) in to (1.9) . This gives:

x = Y 2 ( x−vt ) + Yvt’

From which we find that:

( )
2
1−Y
t’ = x+ Yt…………………………………..(1.12)
Yv

And also from S frame of reference we have a speed of light “c .”

x = ct…………………………………………………(1.13)

And in S’ frame x’ = ct’……………………………………………..(1.14)

Substituting for x’ and t’ in equation (1.14) with the help of equation (1) and equation (1.12) gives:

[ ]
2
1−Y
Y(x – vt) = cx + cYt and solving for x
Yv

[ ( )]
v
cYt +vYt Y+ Y
c
( )c
2
x= 1−Y = ct 2
Y− 1−Y
Yv Y− c
Yv

[ ( )]
v
1+
c
There fore this becomes x = ct ………………………….(1.15)
1 c
1− 2 −1
Y v

The expression for x will be the same as that given by equation (1.13), namely x = ct, provided that the
1
quantity in the bracket equals 1 . There fore using of relativity factor Y =
√ 1−
v 2 we obtain:
c
2

x−vt
x’ =
√ 1−
v 2 ……………………………………………(1.16)
c
2

y’ = y…………………………………………………..(1.17)

z’ = z……………………………………………………(1.18)

3
 vx
t−
c
t’ = …………………………………………….(1.19)

√
2
v
1− 2
c

1.5 Velocity - addition formula

Suppose, relative to a frame S a particle has a velocity : u = uxi^ + uy ^j + uzk^ ………..(1.20)

dx dy dz
Where ux = , uy = and uz =
dt dt dt

Our required is to find the velocity of this particle as measured in the frame of reference S’
moving with the velocity, vx relative to S . If the particle has coordinate at time t in S’ then the
particle will have coordinate x’ at time t’ in S’ where

Vxx '
x = Y( x ' + vxt ' ) and t = Y(t’ + 2
¿
c

If the particle is displaced to a new position x + dx at a time t + dt in S then in S’ it will be at the

position x’ + dx’ at time t’ + dt’ where

x + dx = Y( x ' +d x ' + vx(t ' +d t ' ) )

t + dt = Y( t ' + d t ' +vx (x ' +d x' )/c 2) and hence

dx = Y( d x ' + vxdt ' ) …………………………………(1.21)

dt = Y( dt’ + vxdx’/c 2)………………………………(1.22)

'
dx d x + vxdt '
So that : ux = ⇒ ux = '
dt '
d t +vxd x /c
2

dx '
+ vx
dt ' dx '
ux = where = ux is the velocity of the particle in
vxdx ' dt '
1+ 2
c dt '
the S’ frame of reference. Hence
'
u x + vx
ux = vxux ' ………………………………………………..(1.23)
1+ 2
c

Exercise 1: Using of similar procedure that is y = y’ and z =z’ show that

4
 uy ' uz '
a) uy = & b) uz =
Y ( 1+ vxu x /c )
' 2
Y ( 1+ vxu x ' /c 2)

1.6 Doppler Effect

We are familiar with the increase in pitch of a sound when its source approaches us(or we
approaches the source). These change in frequency constitute the Doppler effect, whose origin
Is straight forward . For instant, successive waves emitted by a source moving to ward an
observer are closer together than normal because of the advance of the source; because the
separation of the waves is the wavelength of the sound, the corresponding frequency is higher.
The relation ship b/n source frequency vo and observed frequency v is :

[ ]
u
1+
c
Doppler effect in sound v = vo ..…………………………………………. (1.24)
v
1−
c

5
Where c = speed of sound

u = speed of observer( + for motion to ward the source, - for motion away from it) .

v = speed of the source( + for motion to wards observer, - for motion away from him) .

If the observer is stationary, u = 0 and if the source is stationary, v = 0. The Doppler effect in
Sound varies depending on which the source, or observer, or both are moving .

1.7 Time Dilation

A clock in moving frame will be seen to be running slow, or “dilated” according to the Lorentz
Transformation . The time will always be shortest as measured in its rest frame. The time
measured in the frame in which the clock is at rest is called the proper time. In the time interval

to = t2’ – t1’ is measured in the moving reference frame. Then t = t2 – t1 can be calculated using
the Lorentz transformation. That is :
' '
' vx 2 ' vx 1
t2+ 2
−t 1 − 2
c c
t = t 2 – t1 =

√
2
v
1− 2
c

The time measurements made in the moving frame are made at the same location, so the
expression reduces to: t = toY and also this can be written as:

√
2
t= v ……………………………………..(1.25)
1− 2
c

Example : A space craft is moving relative to the earth . An observer on the earth finds that, b/n
1 p.m and 2 p.m according to here clock, 3061 s on the space crafts clock. What is the space
craft’s speed relative to the earth ?

Solution

Here to = 3600 s is the proper time interval on the earth and t = 3601 s is the interval in the
moving frame as measured from the earth, we proceed as follows.

√
2
t= v Squaring both sides we have
1− 2
c

6
 ⇒ v = c 1− ¿2 = ( 2.998 m/s ) 1−
√ √
2

( )
2 2
v 2
1-
c ()
¿
2 = t
t
3600 s
3601 s

v = 7.1 x 106 m/s

1.8 Length contraction

The length of any object in a moving frame will appear for shortened in the direction of motion,
or contracted . The amount of contraction can be calculated from the Lorentz transformation.
The length is maximum in the frame in which the object is at rest . If the length L o = x2’ – x1’ is
measured in the moving reference frame, then x 2 – x1 can be calculated using the Lorentz
x 2−vt 2−x 1+ vt 1
Transformation Lo = x2’ - x1’ =
√ v2
1− 2
c

But since the two measurements made the fixed frame are made simultaneously in that frame.

√
t2 = t1 and L = Lo 1−
v2
c
2
=
Lo
Y
…………………………………………………(1.26)

1.9 Relativistic Dynamics

It is a relativistic motion regarding of its cause.

1.9.1 Relativistic Momentum

Any relativistic generalization of Newtonian momentum must satisfy two criteria.

 Relativistic momentum must be conserved in all frame of reference .

 Relativistic momentum must reduce to Newtonian momentum at low speeds.

The relativistic momentum of a particle moving with a velocity u as measured with respect to a
frame of reference S, that satisfies these criteria can be shown to take the form

mou

c
2
√
v 2 ………………………………….(1.27) where mo is the rest of the
P=
1−

Particle. The mass of the particle when at rest, and which can be identified with the Newtonian
mass of the particle. We note immediately that, for u<<c eq (1.27) becomes:

P = mou ……………………………………….(1.28) and it was once that

practice to write the relativistic momentum as

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 P = mu…………………………………………(1.29)

mo
Where m =
√ v 2 = Ymo→ relativistic mass
1− 2
c

1.9.2 Relativistic force, work and energy

dp
Relativistic force is defined as: F = ……………………………………….(1.30)
dt

This force will do work on a particle, and the relativistic work done by F during a small
Displacement dr is, once again defined by analogy as

dw = F.dr ………………………………(1.31)

The rate at which F does work is then P = F.u……………………………(1.32)

Introducing the notation of relativistic kinetic energy by viewing the work done by F as
contributing to wards the kinetic energy of the particle

dT
P= = F.u……………………………………(1.33)
dt

Where T is the relativistic kinetic energy of the particle. We can write this last equation as

dT dp
= F.u = u.
dt dt

(√ )
d mou
dT
= u. dt u2
dt 1− 2
c

(√ )
d u
dT dt
= umo. u2
dt 1−
c2

[√ ]
du
umo+ d 1
dT dt
= + mou.u dt u2

√
2
dt u 1−
1− 2 c2
c

8
 du
umo.

( )
dT dt d u
2 −1/ 2
= + mou . 1− 2
2

√
2
dt u dt c
1− 2
c

du
umo

( )
dT dt 1 u
2 −3 /2
−2 u du
= + mou - 1− 2
2
2 .

( )
2 1 /2
dt u 2 c c dt
1− 2
c

mo
dT du
( )
2 3/2
= u u …………………………(1.34)
dt 1− 2 dt
c

[√ ]
d mo c 2
dT dt
Equation (1.34) is the same as with the expression = u then from this one can
2
dt 1−
c2
write

(√ )
mo c2 2

u ⇒T=
mo c
∫d
∫ dT =
√
u + constant
2 2
1− 1−
c2 c
2

By setting T = u = 0 the constant becomes - moc 2. There fore the total kinetic energy can be
written as: T = Ymoc 2 – moc 2…………………………………………(1.35)

If we suppose that u<<c and using binomial expansion

1
( )
2 −1/ 2 2
u u

√
2
u = 1− 2 =1+ 2
1− 2 c 2c
c

Substituting this in to equation (1.34) we get

2
mou
T= ……………………………… (1.36)
2

A useful relation ship b /n energy and momentum can also be established. Its value lies both in

The starting point is the expression for energy:

9
 2
mo c

( )
2 1 /2
E= u …………………………..(1.37)
1− 2
c

[ ]
2 2
u u
mo c
4
1− 2
+ 2
2 4 c c
u ⇒ E 2 = mo c
2
From which we find E = 2
then the energy becomes
1− 2 u2
c 1− 2
c

E = mo c + P c ⇒ E = √ mo2 c 4 + p 2 c2
2 2 4 2 2

While keeping E and P fixed limit of mo→o

E = Pc

Chapter – Two

2. Development of Quantum Mechanics

Classical mechanics is an approximation of quantum mechanics. The fundamental difference

b/n classical(or Newtonian) mechanics and quantum mechanics lies in what they describe. In
classical mechanics, the future history of particle is completely determined by its initial position
and momentum together with the forces that act up on it. In the every day world these
quantities can all be determined well enough for the predications of Newtonian mechanics a-
gree with what we find.

Quantum mechanics also arrives at relationships b/n observable quantity is different in the
atomic ream. In quantum the kind of certainty about the future characteristic of classical
mechanics of classical mechanics is impossible because the initial state of a particle can not be
established with sufficient accuracy.

2.1 Wave Function

10
The quantity with which quantum mechanics is concerned is the wave function ѱ of the body.
While ѱ it self has no physical interpretation, the square of its of its absolute magnitude │ ѱ │2
evaluated at a particular place at a particular time is proportional to the probability of finding
the body there at that time. The linear momentum, angular momentum, and energy of the
body are quantities that can be established from ѱ.

Wave functions are usually complex both real and imaginary parts. A probability, however,
must be a positive real quantity. The probability density │ ѱ │2 for a complex ѱ is there fore
¿ ¿
taken as the product ѱ ѱ of ѱ and its complex conjugate ѱ .

The complex conjugate of any function is obtained by replacing i( ¿ √ −1 ) by –I where ever it

appears in the function. Every complex function ѱ can be written in the form wave function:

ѱ = A + iB………………………………….(2.1)

Where A and B are real functions. The complex conjugate ѱ ¿ of ѱ is

¿
Complex conjugate ѱ = A- iB………………………………….. (2.2)
¿
And so │ѱ│ = ѱ ѱ = A - i B = A + B
2 2 2 2 2 2

Since i 2 = -1. Hence │ ѱ │2 = ѱ ¿ ѱ is always a positive real quantity, as required.

2.2 Operator

Are mathematical rules in which an old wave function transforms in to new wave function. An
operator is represented by a carat over its head. If ^
A is an operator, ^
A ѱ(x,t) = ѱ(x,t). Every
dynamical variable in classical mechanics is represented by an operator in a quantum
mechanics.

Dynamical Variable Operators

X,y,z ^x , ^y , ^z

r r^

f(x,y,z) ^f (x,y,z)

V(r) ^v (r)

E ^
E

Example: ѱ(x,t) = A e i (kx−ωt )

11
 ∂ ∂
Let us derive both sides by and
∂t ∂x

∂ ∂
ѱ(x,t) = ( A e i (kx−ωt ) )
∂t ∂t

∂
ѱ(x,t) = -iω( A e (kx −ωt ) )
∂t

By multiplying both sides by –i ţ

𝜓(x,t) = - i 2ţωAe i (kx−ωt )

∂
iţ
∂t

𝜓(x,t) = ^
E 𝜓(x,t)
∂
But ţω = ^
E then iţ
∂t

^ ∂
E = iţ
∂t

∂
And also to find momentum let’s derivate both sides by . i.e
∂x

𝜓(x,t) =
∂ ∂
( A ei ( kx−ωt ))
∂x ∂x

𝜓(x,t) = ikAe i (kx−ωt )

∂
∂x

Multiplying both sides by –iţ we get

𝜓(x,t) = -i 2 ţ kAe i (kx−ωt )

∂
-iţ
∂x

𝜓(x,t) = ^
Px𝜓(x,t) since ^
∂
-iţ P = ţk
∂x

^ ∂
Therefore Px = -iţ
∂x

^ ∂ ^ ∂
Similarly Py = -iţ , Pz = -iţ
∂y ∂z

P zk^ ⇒ ^
^ ∂ ^ ∂ ^ ∂
P x i^ + ^
P=^ P y ^j + ^ P = -iţ i - iţ j - iţ ^z
∂x ∂y ∂z

12
By taking common factor –iţ we have ^
P = -iţ ( ∂∂x i^ + ∂∂y ^j+ ∂∂z k^ )
^
P = -iţ∇

∂ ^ ∂ ^ ∂
Where ∇= i+ j+ ^z
∂x ∂y ∂z

2.2.1 Hamiltonian Operator

They are operators which represents the total energy.

^
H=^
E →Schrodinger equation

^2
P
^
H = T^ + ^v (r) = + ^v (r)
2m
2
^ (−iţ ∇ )
H= + ^v (r)
2m

Where ^v → is potential energy operator

T^ → is kinetic energy operator

2.3 Normalization
∞

Normalization condition can be given by ∫ │ ѱ │2 dv = 1…………………..(2.3)

−∞

Since if the particle exists some where at all times. ∫ Pdv = 1

−∞

The wave function that obeys equation (2.3) is said to be normalized. Every acceptable wave
function can be normalized by multiply it by an appropriate constant.

2.3.1 Well behaved wave function

A function is said to be well behaved wave function if and only if the following condition holds.

 Ѱ must be continues and single valued every where.

∂ѱ ∂ѱ ∂ѱ
 , and must be continues and single valued every where.
∂x ∂ y ∂z
 Ѱ must be normalizable, which means that ѱ must be go to zero as x→ ± ∞ , y→ ± ∞, z
→ ± ∞ in order that ∫ │ ѱ │ dv over all space be a finite constant.
2

13
For a particle restricted to motion in the x- direction, the probability of finding it b/n x 1 and x2 is
given by:
x2

Px1x2 =∫ │ ѱ │ dv……………………………..(2.4)
2
Probability
x1

2.4 Expectation Value

The expectation value of the position of the single particle is:

∞

∫ x │ѱ │2
−∞
¿ x >¿ = ∞ dx………………………….(2.5)
∫ │ѱ │ 2

−∞

If ѱ is normalized wave function, the denominator of equation (2.5) equals the probability that
the particle exists somewhere b/n x = -∞ and x = ∞ and there fore has the value 1. In this case
the expectation value for position is:
∞
¿ x >¿ = ∫ x │ѱ │2dx…………………………………..(2.6)
−∞

Example: A particle limited to the x-axis has the wave function ѱ = ax b/n x =0 and x = 1; ѱ = 0
else where. Then

a) Find the probability that the particle can be found b /n x = 0.45 and x = 0.55.
b) Find the expectation value of the particle position.
Solution

[ ]
x2 x2 3
x 0.55
a) The probability is: ∫ │ ѱ │ dx = a 2 ∫ x dx = a 2
2 2

3
│0.45 = 0.0251a 2
x1 x1

1 1 2
a
b) The expectation value is : ¿ x >¿ = ∫ x │ѱ │ dx = a ∫ x dx =
2 2 3

0 0 4

2.5 Eigen value and Eigen function

The value of energy En for which Schrodiger’s stead state equation can be solved are called
eigen values and the corresponding wave functions ѱ n are called eigen functions. In short

^ ѱ n = Gn ѱ n ……………………………….(2.7)
G

14
Where Gn → is eigen value

ѱ n → is eigen fuction
2
d 2x
Example: An eigen function of the operator 2 is ѱ = e . Then find the corresponding eigen
dx
value of the operator.
2
^ n = d , so using equation (2.7) we have:
Hence G 2
dx
2
^ = d 2 ( e 2 x ) = 4e 2 x ⇒ G
Gѱ ^ nѱ = 4ѱ
dx

From this we see that the eigen value G here is just G = 4.

2.6 Particle in a box

In quantum mechanics, the particle in a box model describes a particle free to move in small
space surrounded by impenetrable barriers, likewise it can never have zero energy. We may
specify the particles motion by saying that it is restricted to traveling along the x axis b/n x =0
and x = L by infinitely hard walls. From formal point of view the potential energy ‘’u‘’of the
particle is infinite on both sides of the box , while u is the constant say zero for convenience on
the in side as shown below. Because the particle can not have an infinite amount of energy, it
can not exist out side the box, and so its wave function ѱ is zero for x≤ 0 and x≥ L. Our task is
to find what ѱ is with in the box Schrodinger equation becomes:
2
d ѱ 2mEѱ
2 + 2 = 0………………(2.8)
dx ţ

F.g (2.1) A square potential wall with infinity high barriers at each end corresponding to a box
with infinity hard walls.

15
 2 2
d ѱ ∂ ѱ
Since the total derivative 2 is the same as the partial derivative 2 because ѱ is a
dx ∂x
function only of x in this problem. So, the solution of equation (2.8) is:

ѱ = Asin
√2 mE x + Bcos √2 mE x……………..(2.9)
ţ ţ

This solution is subject to the boundary conditions that ѱ = 0 f0r x = 0and for x = L. since cos0 =
1, the second term can not describe the particle because it does not vanish at x = 0. Hence we
conclude that B = 0,the sin term always yields ѱ = 0 at x = 0, s required, but ѱ will be zero at x =
L when

√2 mE L = n π … … … … ……………………..(2.10)
ţ

where n = 1,2,3…

This result comes about the sins of the angles π , 2 π , 3 π ….are all zero. From equation (2.10) it
is clear that the energy of the particle can have only certain values.
2 2 2
n π ţ
Particle in a box En = 2 ………………………………………….(2.11) for n = 1,2,3
2 ml

The wave function of the particle in a box whose energies are En are

Asinnπx
ѱn = ……………………………………….(2.12)
L

Then the normalized wave function of equation (2.11) is :

∞ l l

∫ │ ѱ n │ dx = ∫ │ ѱ n │ dx = A
2 2 2
∫ sin2 ( nπx
l )
dx
−∞ 0 0

1
Since sin2 θ = ( 1−cos 2θ ) using this identity we have
2

[ ]
∞ l l

( )
2 2
A 2nπx A L
∫ │ ѱ n │ dx =
2

2
∫ dx−∫ cos l
dx =
2
−∞ 0 0

= 1 ⇒ A2 ( 2l ) = 1.
∞

And ∫ │ ѱ n │2dx
−∞

16
 A=
√ 2
L

The normalized wave function of the particle are :

Particle in the box

√
ѱ n = 2 sin
l
nπx
L

Excericse: Calculate the normalized wave function of a particle in a box whose wave function
are given by:

Acosnπx
ѱn =
L

Chapter – Three

3. Wave Properties of Particles

17